Prove using the definition on the limit that \lim_{n\to\infty}\frac{1+n}{1-2n}=\frac{-1}{2}

slaggingV 2021-08-19 Answered
Prove using the definition on the limit that
\(\displaystyle\lim_{{{n}\to\infty}}{\frac{{{1}+{n}}}{{{1}-{2}{n}}}}={\frac{{-{1}}}{{{2}}}}\)

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Expert Answer

wheezym
Answered 2021-08-20 Author has 13806 answers

\(\Bigg|\frac{1+n}{1-2n}+\frac{1}{2}\Bigg|\)
\(=\Bigg|\frac{2+2n+1-2n}{2(1-2n)}\Bigg|\)
\(=\Bigg|\frac{3}{2}\cdot\frac{1}{1-2n}\Bigg|\)
For every \(\displaystyle\epsilon{>}{0}\) given we can choose N that
\(\Bigg|\frac{3}{2}\cdot\frac{1}{1-2}=\frac{3}{2(2N-1)}\Bigg|<\epsilon\)

\(\displaystyle{2}{N}-{1}{>}{\frac{{{3}}}{{{2}\epsilon}}}\)
\(\displaystyle{N}{>}{\left({1}+{\frac{{{3}}}{{{2}\epsilon}}}\right)}{\frac{{{1}}}{{{2}}}}\)
\(\displaystyle{N}{>}{\left({\frac{{{1}}}{{{2}}}}+{\frac{{{3}}}{{{4}\epsilon}}}\right)}\)
Hence \(\displaystyle\forall{n}\geq{N}\)
\(\Bigg|\frac{1+2n}{1-2n}-\bigg(-\frac{1}{2}\bigg)\Bigg|<\epsilon\)
Hence
\(\lim_{n\to\infty}\Bigg|\frac{1+2n}{1-2n}\Bigg|=-\frac{1}{2}\)

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